For the second part I see now that for $g \in V^*$ and $f \in W^*$ we must have $T^*(f) = g$. Let $v_1,\dots,v_n$ be a basis of $V$; then the list $Tv_1,\dots,Tv_n$ is linearly independent in $W$, so we can extend it to a basis $Tv_1,\dots,Tv_n,w_1,\dots,w_k$ of $W$. Now we can define $f : W \rightarrow F$ by setting $f(Tv_i) = g(vi)$$f(Tv_i) = g(v_i)$ and $f(w_i) = 0$. Then $T^*(f) : V \rightarrow F$ and $T^*(f)(v_i) = (f \circ T)(v_i) = f(Tv_i) = g(v_i)$, so $T^*(f) = g$, as desired.
For the second part I see now that for $g \in V^*$ and $f \in W^*$ we must have $T^*(f) = g$. Let $v_1,\dots,v_n$ be a basis of $V$; then the list $Tv_1,\dots,Tv_n$ is linearly independent in $W$, so we can extend it to a basis $Tv_1,\dots,Tv_n,w_1,\dots,w_k$ of $W$. Now we can define $f : W \rightarrow F$ by setting $f(Tv_i) = g(vi)$ and $f(w_i) = 0$. Then $T^*(f) : V \rightarrow F$ and $T^*(f)(v_i) = (f \circ T)(v_i) = f(Tv_i) = g(v_i)$, so $T^*(f) = g$, as desired.
For the second part I see now that for $g \in V^*$ and $f \in W^*$ we must have $T^*(f) = g$. Let $v_1,\dots,v_n$ be a basis of $V$; then the list $Tv_1,\dots,Tv_n$ is linearly independent in $W$, so we can extend it to a basis $Tv_1,\dots,Tv_n,w_1,\dots,w_k$ of $W$. Now we can define $f : W \rightarrow F$ by setting $f(Tv_i) = g(v_i)$ and $f(w_i) = 0$. Then $T^*(f) : V \rightarrow F$ and $T^*(f)(v_i) = (f \circ T)(v_i) = f(Tv_i) = g(v_i)$, so $T^*(f) = g$, as desired.
For the second part I see now that for $g \in V^*$ and $f \in W^*$ we must have $T^*(f) = g$. Let $v_1,\dots,v_n$ be a basis of $V$; then the list $Tv_1,\dots,Tv_n$ is linearly independent in $W$, so we can extend it to a basis $Tv_1,\dots,Tv_n,w_1,\dots,w_k$ of $W$. Now we can define $f : W \rightarrow F$ by setting $f(Tv_i) = g(vi)$ and $f(w_i) = 0$. Then $T^*(f) : V \rightarrow F$ and $T^*(f)(v_i) = (f \circ T)(v_i) = f(Tv_i) = g(v_i)$, so $T^*(f) = g$, as desired.